Length Contraction rearrangement

In summary, In order to find the length of an object in a given frame of reference, the time coordinates must have the same value. Suppose the object is in rest in another frame of reference. Then, you can use the equation 1) to calculate the length contraction, or 2) use the reverse Lorentz transformation to impose Δt=0 and get x'_2 - x'_1 = \gamma (x_2 - x_1).
  • #1
mananvpanchal
215
0
Hello,

Suppose, There is two points [itex][t_{a1}, x_{a1}][/itex] and [itex][t_{a2}, x_{a2}][/itex] stationary in my frame A. I say [itex]L_a=x_{a2}-x_{a1}[/itex].

If I want to find x component of co-ordinates in other frame B which is moving relative to me with constant speed v.

I have to use the equation

1. [itex]x_b=\gamma(x_a-vt_a)[/itex], where [itex]\gamma > 1[/itex].

So, I can get [itex]x_{b1}=\gamma(x_{a1}-vt_{a1})[/itex], [itex]x_{b2}=\gamma(x_{a2}-vt_{a2})[/itex].

If I take [itex]t_{a1}=t_{a2}=0[/itex], then [itex]x_{b1}=\gamma x_{a1}[/itex] and [itex]x_{b2}=\gamma x_{a2}[/itex].

Now, how much distance B measures in his co-ordinate system between this two points is [itex]x_{b2} - x_{b1} = \gamma (x_{a2} - x_{a1})[/itex].

If B says [itex]L_b=x_{b2} - x_{b1}[/itex].

So, I would get [itex]L_b=\gamma L_a[/itex].

Here, [itex]L_b > L_a[/itex]. But, this should not be the case.

If I want length contraction then I have to derive it oppositely. (I found this method from http://en.wikipedia.org/wiki/Length_contraction#Derivation and http://www.fourmilab.ch/etexts/einstein/specrel/www/#SECTION14)

I have to pick some other B's co-ordinates [itex][t_{b1}, x_{b1}][/itex] and [itex][t_{b2}, x_{b2}][/itex] which is stationary in B's frame. B says [itex]L_b=x_{b2}-x_{b1}[/itex]

I have to use this equation

2. [itex]x_{a}=\gamma(x_{b}-vt_{b})[/itex], where [itex]\gamma > 1[/itex].

I would find [itex]x_{a1}=\gamma(x_{b1}-vt_{b1})[/itex] and [itex]x_{a2}=\gamma(x_{b2}-vt_{b2})[/itex].

If I take [itex]t_{b1}=t_{b2}=0[/itex] then, [itex]x_{a1}=\gamma x_{b1}[/itex] and [itex]x_{a2}=\gamma x_{b2}[/itex].

So, [itex]x_{a2} - x_{a1} = \gamma (x_{b2} - x_{b1})[/itex].

I says [itex]L_a=x_{a2}-x_{a1}[/itex].

so, [itex]L_a=\gamma L_b[/itex].
so, [tex]L_b=\frac{L_a}{\gamma}[/tex]
so, [itex]L_b < L_a[/itex]. Now, we can say this as length contraction.

But, I have started this derivation using B's co-ordinates. I as A don't know B's co-ordinates. I only know my co-ordinates because I can physically define it. I have to calculate B's co-ordinates to get Length Contraction.

I cannot use equation (1) for that, it wouldn't give me length contraction.

How can I get B's co-ordinates using my own?

If I as A has a stationary point [itex][t_a, x_a][/itex], I can calculate B's co-ordinate using my own in Galilean transformation.
 
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  • #2
Whenever you want to determine the length of an object in a given Frame of Reference, the time coordinates must have the same value. You fulfilled this requirement in the first frame when you said ta1=ta2=0 but when you transform the two events at either end of the object into the second frame, the new time coordinates will not be equal, tb1≠tb2.

One way to find a pair of events that will meet the requirement when the first event is at the origin in both frames is to multiply the x-component of the second event in the first frame by the velocity (assuming c=1) and use that as the time-coordinate for the second event in the first frame. Now when you perform the Lorentz Transform on the two events, the time coordinates will be equal and you will get the correct contracted length.

If the first event is not at the origin, then you have to take the delta between the x-components and multiply that by the velocity and add that to the time-coordinate of the first event and use that as the time-coordinate of the second event.

So, in your example, let's make the added assumption that xa1=0 which will put the first event at the origin. The we multiply xa2 by v to get vxa2 so the second event is [vxa2,xa2].

When you transform the time-coordinate for this event you get tb2=γ(vxa2-vxa2)=0, which is what you want, the t coordinate for the second transformed event equal to the first transformed event (which is at the origin).

When you transform the x-coordinate for this event you get xb2=γ(xa2-vvxa2)=γ(xa2)(1-v2) and since γ=1/√(1-v2), this simplifies to xb2=x√(1-v2)=x/γ, which is what you want.
 
  • #3
Ok, please see the wiki http://en.wikipedia.org/wiki/Length_contraction#Derivation.

Object is in rest in [itex]S'[/itex] frame, so there is always [itex]t'_1=t'_2[/itex].
But, the article says we have to put [itex]t_1=t_2[/itex] to calculate length [itex]L[/itex]. How is that possible?

As I understand we cannot get [itex][t'_1, x'_1][/itex] and [itex][t'_2, x'_2][/itex] for any [itex][t_1, x_1][/itex] and [itex][t_2, x_2][/itex] where [itex]t'_1=t'_2[/itex] and [itex]t_1=t_2[/itex].

Please, let me know if we would get this.

Suppose, the article not saying that [itex]t_1=t_2[/itex] is a requirement, then we cannot get

[itex]x'_2 - x'_1 = \gamma (x_2 - x_1)[/itex].
 
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  • #4
mananvpanchal said:
Object is in rest in [itex]S'[/itex] frame, so there is always [itex]t'_1=t'_2[/itex].
Not really. Since the object is at rest in S', the ends can be measured at different times. What's important is that Δx' = L0 (the rest length of the object).
But, the article says we have to put [itex]t_1=t_2[/itex] to calculate length [itex]L[/itex]. How is that possible?
In order for S to measure the length of the moving object, he must measure the positions of the ends at the same time (according to S). Thus Δt = 0 must be imposed.

Using the reverse Lorentz transformation [itex]\Delta x' = \gamma(\Delta x - v\Delta t)[/itex] is all you need to get the length contraction formula.
 
  • #5
mananvpanchal said:
Ok, please see the wiki http://en.wikipedia.org/wiki/Length_contraction#Derivation.

Object is in rest in [itex]S'[/itex] frame, so there is always [itex]t'_1=t'_2[/itex].
But, the article says we have to put [itex]t_1=t_2[/itex] to calculate length [itex]L[/itex]. How is that possible?

As I understand we cannot get [itex][t'_1, x'_1][/itex] and [itex][t'_2, x'_2][/itex] for any [itex][t_1, x_1][/itex] and [itex][t_2, x_2][/itex] where [itex]t'_1=t'_2[/itex] and [itex]t_1=t_2[/itex].

Please, let me know if we would get this.

Suppose, the article not saying that [itex]t_1=t_2[/itex] is a requirement, then we cannot get

[itex]x'_2 - x'_1 = \gamma (x_2 - x_1)[/itex].
I don't understand why you pointed me to another wikipedia article that says the same thing I said which is that the time coordinates for the two events that define the ends of the object for which you want to know its length in a given Frame of Reference must be equal. The article doesn't give you a clue as to how you arrive at such a pair of events which is what I was giving you. Did you not understand my method?
 
  • #6
ghwellsjr said:
Whenever you want to determine the length of an object in a given Frame of Reference, the time coordinates must have the same value. You fulfilled this requirement in the first frame when you said ta1=ta2=0 but when you transform the two events at either end of the object into the second frame, the new time coordinates will not be equal, tb1≠tb2.

One way to find a pair of events that will meet the requirement when the first event is at the origin in both frames is to multiply the x-component of the second event in the first frame by the velocity (assuming c=1) and use that as the time-coordinate for the second event in the first frame. Now when you perform the Lorentz Transform on the two events, the time coordinates will be equal and you will get the correct contracted length.

If the first event is not at the origin, then you have to take the delta between the x-components and multiply that by the velocity and add that to the time-coordinate of the first event and use that as the time-coordinate of the second event.

So, in your example, let's make the added assumption that xa1=0 which will put the first event at the origin. The we multiply xa2 by v to get vxa2 so the second event is [vxa2,xa2].

When you transform the time-coordinate for this event you get tb2=γ(vxa2-vxa2)=0, which is what you want, the t coordinate for the second transformed event equal to the first transformed event (which is at the origin).

When you transform the x-coordinate for this event you get xb2=γ(xa2-vvxa2)=γ(xa2)(1-v2) and since γ=1/√(1-v2), this simplifies to xb2=x√(1-v2)=x/γ, which is what you want.

Yes, I understand what you are saying. We want [itex]t_{b1}=t_{b2}[/itex]. So, for that we made [itex][t_{a1}, x_{a1}]=[0, 0][/itex] and we found [itex]t_{a2}=t_{a1}+v(x_{a2}-x_{a1})=vx_{a2}[/itex]. Now we transform, we get [itex]t_{b1}=0[/itex] and [itex]t_{b2}=0[/itex]. So the [itex](x_{b2}-x_{b1})[/itex] is proper length in B frame, because we have [itex]t_{b1}=t_{b2}[/itex]. But we have [itex]t_{a1}{\neq}t_{a2}[/itex] now.
This is what I am trying to say. If you make time co-ordinate same in one frame, you never get same time co-ordinate in transformed frame. So if we do [itex](x_{a2}-x_{a1})[/itex], does this gives us exact contracted length?

Other thing I want to point out is in wiki
wiki said:
In an inertial reference frame S', [itex]x'_1[/itex] and [itex]x'_2[/itex] shall denote the endpoints for an object of length [itex]L'_0[/itex] at rest in this system.
So, [itex]L'_0[/itex] at rest in S', and so [itex]L'_0[/itex] is proper length in S', and so we should have [itex]t'_1=t'_2[/itex].

And then wiki says
wiki said:
we have to put [itex]t_1=t_2[/itex]

Is this possible?
 
  • #7
mananvpanchal said:
Other thing I want to point out is in wiki
wiki said:
In an inertial reference frame S', [itex]x'_1[/itex] and [itex]x'_2[/itex] shall denote the endpoints for an object of length [itex]L'_0[/itex] at rest in this system.
So, [itex]L'_0[/itex] at rest in S', and so [itex]L'_0[/itex] is proper length in S', and so we should have [itex]t'_1=t'_2[/itex].
Where in the Wiki article does it say that t'1 = t'2?
 
  • #8
Doc Al said:
Not really. Since the object is at rest in S', the ends can be measured at different times. What's important is that Δx' = L0 (the rest length of the object).

Yes, we can measure at different time, but that doesn't mean two ends has different time value.
This is not the measurement time. When a object is in rest, then the object has same time component value. But when the object is moving relative to a S frame then the object has different time component value for S. (Relativity Of Simultaneity)

When some event happen near you at same time and a event happen far from you. But, you see the near event first, and after that the far event. That doesn't mean the event is not simultaneous.

Please, read "What the Relativity of Simultaneity is NOT" in the link
http://www.pitt.edu/~jdnorton/teaching/HPS_0410/chapters/Special_relativity_rel_sim/
 
  • #9
mananvpanchal said:
Yes, I understand what you are saying. We want [itex]t_{b1}=t_{b2}[/itex]. So, for that we made [itex][t_{a1}, x_{a1}]=[0, 0][/itex] and we found [itex]t_{a2}=t_{a1}+v(x_{a2}-x_{a1})=vx_{a2}[/itex]. Now we transform, we get [itex]t_{b1}=0[/itex] and [itex]t_{b2}=0[/itex]. So the [itex](x_{b2}-x_{b1})[/itex] is proper length in B frame, because we have [itex]t_{b1}=t_{b2}[/itex]. But we have [itex]t_{a1}{\neq}t_{a2}[/itex] now.
This is what I am trying to say. If you make time co-ordinate same in one frame, you never get same time co-ordinate in transformed frame.
Yes, that is in general true. That's the problem. The length of an object must be determined with the two events at its endpoints taken at the same time which means when we transform those two events into another frame, they almost always won't have the same time components. But we can't just change the time of one of the transformed events to make it equal to the time of the other transformed event, we have to change the time of one (or both) of the events in the first frame so that when we transform the two events into the second frame, the new time coordinates come out the same. It doesn't matter how we do this, I just gave you one way, you could also do it by trial and error.
mananvpanchal said:
So if we do [itex](x_{a2}-x_{a1})[/itex], does this gives us exact contracted length?
You indicated in your first post that when you just transform the end-point events from the rest frame to the moving frame, you get a length that is multiplied by gamma instead of divided by gamma, correct? I showed you how to get a length that is divided by gamma which we already know is the correct answer.
mananvpanchal said:
Other thing I want to point out is in wiki
wiki said:
In an inertial reference frame S', x′1 and x′2 shall denote the endpoints for an object of length L′0 at rest in this system.
So, [itex]L'_0[/itex] at rest in S', and so [itex]L'_0[/itex] is proper length in S', and so we should have [itex]t'_1=t'_2[/itex].

And then wiki says
wiki said:
we have to put t1=t2
Is this possible?
Here they are saying that in S', the object is at rest and when you want to know its length in S, you have to find a pair of events in S' that when transformed into S give t1=t2, but they don't tell you how to do that. It is possible by any number of ways.

If you want, you can cheat by knowing that the moving length should be the rest length divided by gamma and picking a pair of events in the second frame separated by that length and with the times equal, and then transforming those events into the first frame and just changing the times so they are equal but now the length will be gamma times the length in the first frame and you can pretend that you now want to transform that new event to the second frame which, of course will give you another length that is too large, but then you can say, aha, I know a set of times in the first frame that will transform into the second frame and result in two events with the same time on them and then it will give the correct length. This might be a good way to fool someone else when you're showing them how you need to select two times in the first frame that are not equal so that the two times in the second frame will be equal, but like I say, it's cheating because you worked the problem out backwards.

So another way is by trial and error. All you have to do is start with the two end-point events with simultaneous time coordinates in the first frame, transform them to see that the new events are not simultaneous, then tweak either one of the old time coordinates, do the transformations again, see if the new time coordinates got closer together and if so, continue tweaking the same old time coordinate in the same direction until the new time coordinates match. If the new time coordinates get farther apart, go back and tweak the old time coordinate in the other direction and continue until you find a pair of old events that transform into new events that are simultaneous.

Or you could do my way.

Just remember the important point is that when you are changing the time coordinates of the old events, you must not change the spatial coordinates because that would be determining a different length.
 
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  • #10
mananvpanchal said:
Yes, we can measure at different time, but that doesn't mean two ends has different time value.
This is not the measurement time. When a object is in rest, then the object has same time component value. But when the object is moving relative to a S frame then the object has different time component value for S. (Relativity Of Simultaneity)
You need to ask yourself what x,t and x',t' represent in the context of deriving the length contraction formula. An observer measures the location of the endpoints of a moving object. The coordinates of those measurements are x1,t1 and x2,t2. Of course, t1=t2. But to observers in S', those measurements were not made at the same time: t'1≠t'2.
 
  • #11
mananvpanchal said:
So, L′0 at rest in S', and so L′0 is proper length in S', and so we should have t′1=t′2.

No, the simultaneous measurement of the endpoints is only necessary in S, the frame in which the rod is in relative motion. Therefore one simply states: t1=t2.

Using the reverse Lorentz transformation, one obtains t'1≠t'2 for the rod's rest frame S'. But this doesn't matter because in S' you can measure the endpoints whenever you want, since the rod is at rest here.

Regards,
 
  • #12
There's confusion going on here because when we are talking about events in a Frame of Reference and using the Lorentz Transform to get the coordinates of those events in another FoR, there is no consideration for any measurement taking place and no observer located in the scenario to make any measurement, rather, it is pure calculation.

It is true that when a real observer is making a real measurement with a real ruler of a real object that is at rest with respect to him, that the time over which he makes the measurement is of no consequence and, in fact, there does not need to be any FoR assigned or considered. Making measurements is not an issue with a FoR.

Furthermore, a real observer making a real measurement of a real object in motion does not need to do it in an instant of time. He can first measure its speed and then he can measure how long it takes for the object to pass by a particular location and then he can calculate its length.

But this thread is not about how an observer measures an object but rather how to correctly transform events from one frame to another in a meaningful way, so let's not confuse the two different ideas.
 
  • #13
ghwellsjr said:
So another way is by trial and error. All you have to do is start with the two end-point events with simultaneous time coordinates in the first frame, transform them to see that the new events are not simultaneous, then tweak either one of the old time coordinates, do the transformations again, see if the new time coordinates got closer together and if so, continue tweaking the same old time coordinate in the same direction until the new time coordinates match. If the new time coordinates get farther apart, go back and tweak the old time coordinate in the other direction and continue until you find a pair of old events that transform into new events that are simultaneous.

Or you could do my way.

Just remember the important point is that when you are changing the time coordinates of the old events, you must not change the spatial coordinates because that would be determining a different length.

Its look like, we have bunch of time components for the rest rod in our frame. And we can pick any pairs which can give us equal time components in transformed frame.

But, this is not possible. You can have only one time component for all space components of rod if rod is at rest in your frame.
Suppose, a long object is in your frame, and If you put two synchronized clocks at both ends. After any how much time you find the clocks is still synchronized. You cannot get different time components for both ends. Actually you have different space component at different different points of rod, but the space components is surely constant with time, it is not changing with time. When you start moving the long object in your reference frame, now time component of both ends surely changed and you get clocks no longer synchronized.

How can you draw a Minkowsky diagram for rest rod with different time components? If you draw a rest rod, it looks like horizontal line parallel to space axis. It has one time component for all space component values. If you change the time component then there is no longer horizontal line, now it look like diagonal line. Which is same as diagram of moving rod in your reference frame.
 
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  • #14
Lets talk about how you would go about measuring the length of an object that is at rest in the S' frame of reference when you are an observer (or, more precisely, a member of a team of observers) at rest in the S frame of reference. The other members of your team are strung out along the direction of motion x. In the S' frame of reference, the object is always situated between coordinates x' and x'+L', with the trailing edge at x' and the leading edge at x'+L'. One way of measuring the length of the object, as reckoned from your frame of reference, is for the various members of your team to each record on a sheet of paper the times displayed on each of their (synchronized) clocks at which the leading edge and the trailing edge of the object pass their locations, together with their own coordinates x. They then get together afterwards and make a plot of x on the vertical axis vs t on the horizontal axis both for the leading edge and the trailing edge of the object (on the same piece of graph paper). How can they then determine how long the object is from their perspective in the S reference frame? The only way that makes sense is to determine the vertical distance L between the two parallel lines on the graph. But this is the distance at each constant value of the time t as measured on the clocks in your frame of reference. This is why you use constant t in the Lorentz Transformation calculations.
 
  • #16
mananvpanchal said:
Its look like, we have bunch of time components for the rest rod in our frame. And we can pick any pairs which can give us equal time components in transformed frame.

But, this is not possible. You can have only one time component for all space components of rod if rod is at rest in your frame.
I tried to warn you in post #12 that there are two different things going on here. You started out asking about how to use the Lorentz Transform to calculate the length of an object in a frame moving with respect to its rest frame. When you do this, we don't care about the existence of any observers making any measurements. It's pure calculation. When we talk about a thought problem identified within the context of a Frame of Reference, we are free to consider any event at any location in any order we please. It's just math.

When you say that we can have only one time component for all space components of a rod, how would you respond if I said, Well that violates the principle that you can't go faster than the speed of light? How do you get from one end of the rod to the other end of the rod at the same instant in time?

Obviously, no one is traversing that distance, we are merely assigning coordinates to those different positions and showing that the length of the rod that we previously measured over a period of time is the same as what the spatial coordinates of the events indicate. In other words, we previously measured the length of the rod by placing a ruler next to the rod and actually moving from one end of the rod to line up one end of the ruler and then moving to the other end of the rod and looking to see what the markings on the ruler indicate its length to be. This cannot be done in an instant of time because we cannot travel instantly from one end of the rod to the other end.

So I don't see why you are saying you can only have one time component for all space components.
mananvpanchal said:
Suppose, a long object is in your frame, and If you put two synchronized clocks at both ends. After any how much time you find the clocks is still synchronized. You cannot get different time components for both ends. Actually you have different space component at different different points of rod, but the space components is surely constant with time, it is not changing with time.
But since this is true, it is also true that you can pick any event at one end of the rod and any other event at the other end of the rod, such that the two events have different time components, and if you subtract the two space components, you will get to correct length of the rod, correct?
mananvpanchal said:
When you start moving the long object in your reference frame, now time component of both ends surely changed and you get clocks no longer synchronized.
That's because the length of the rod has changed and the two clocks were not accelerated identically. What's your point?
mananvpanchal said:
How can you draw a Minkowsky diagram for rest rod with different time components? If you draw a rest rod, it looks like horizontal line parallel to space axis. It has one time component for all space component values. If you change the time component then there is no longer horizontal line, now it look like diagonal line. Which is same as diagram of moving rod in your reference frame.
If you wanted to measure the length of the long rod, moving with respect to you, and you previously had set up some synchronized clocks at measured locations that remain at rest with respect to you, you could observe the progress of the rod as it passed these various clocks at known locations (after the fact, of course, because it takes time for the information to get to you) and from this you could determine its speed. Then you could see how long it took for the object to traverse a given point and then you could calculate its length, correct?

Now after having "measured" the length of the rod in this manner, you can also just look at the record of where the two ends were at every point in time and pick a pair of coordinates where the times were the same and subtract the end positions and you will get the same length as you "measured" earlier. This is essentially what Chestermiller described for you in post #14.

EDIT: This last process is essentially the trial and error method I mentioned earlier.
 
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  • #17
ghwellsjr said:
mananvpanchal said:
Its look like, we have bunch of time components for the rest rod in our frame. And we can pick any pairs which can give us equal time components in transformed frame.

But, this is not possible. You can have only one time component for all space components of rod if rod is at rest in your frame.
I tried to warn you in post #12 that there are two different things going on here. You started out asking about how to use the Lorentz Transform to calculate the length of an object in a frame moving with respect to its rest frame. When you do this, we don't care about the existence of any observers making any measurements. It's pure calculation. When we talk about a thought problem identified within the context of a Frame of Reference, we are free to consider any event at any location in any order we please. It's just math.

Why do you warn me? I am not talking about measurement. I am also curious about just calculation.

Again I already got the point that to calculate x component we have to take different time component at both end. I was just trying to say that how rest object can have two time component. If I am agree on this then I have another question.

ghwellsjr said:
When you say that we can have only one time component for all space components of a rod, how would you respond if I said, Well that violates the principle that you can't go faster than the speed of light? How do you get from one end of the rod to the other end of the rod at the same instant in time?

If there are two simultaneous events, how can we determine the simultaneity by perceiving the events simultaneously or unsimultaneously?
 
  • #18
mananvpanchal said:
If there are two simultaneous events, how can we determine the simultaneity by perceiving the events simultaneously or unsimultaneously?
An event in Special Relativity is one coordinate of time (a number) and three coordinates of space (three numbers). If you haven't specified your scenario enough, meaning you haven't assigned numbers to your events or there is no way to calculate them, then you can't determine if any pair of events are simultaneous or not. If you have, then you just look at the values of the time coordinates for the two events in question and if they are equal, the two events are simultaneous.

I bet I haven't answered your question. If I haven't, you'll have to rephrase it because it really doesn't make much sense to me.
 
  • #19
ghwellsjr said:
An event in Special Relativity is one coordinate of time (a number) and three coordinates of space (three numbers). If you haven't specified your scenario enough, meaning you haven't assigned numbers to your events or there is no way to calculate them, then you can't determine if any pair of events are simultaneous or not. If you have, then you just look at the values of the time coordinates for the two events in question and if they are equal, the two events are simultaneous.

I bet I haven't answered your question. If I haven't, you'll have to rephrase it because it really doesn't make much sense to me.

simult_events.JPG


Please, look at the events you can give any space co-ordinates to event A like [itex](t, x_a, 0, 0)[/itex] and to event B like [itex](t, x_b, 0, 0)[/itex]. We can see time co-ordinate is same for both. But, the events perceives to Observer at middle at different time like [itex]t_a[/itex] and [itex]t_b[/itex].

How can observer determine space components and time component of the events?
 
  • #20
mananvpanchal said:
View attachment 45024

Please, look at the events you can give any space co-ordinates to event A like [itex](t, x_a, 0, 0)[/itex] and to event B like [itex](t, x_b, 0, 0)[/itex]. We can see time co-ordinate is same for both. But, the events perceives to Observer at middle at different time like [itex]t_a[/itex] and [itex]t_b[/itex].

How can observer determine space components and time component of the events?
You need to re-read what I wrote in post #12:
ghwellsjr said:
There's confusion going on here because when we are talking about events in a Frame of Reference and using the Lorentz Transform to get the coordinates of those events in another FoR, there is no consideration for any measurement taking place and no observer located in the scenario to make any measurement, rather, it is pure calculation.

It is true that when a real observer is making a real measurement with a real ruler of a real object that is at rest with respect to him, that the time over which he makes the measurement is of no consequence and, in fact, there does not need to be any FoR assigned or considered. Making measurements is not an issue with a FoR.

Furthermore, a real observer making a real measurement of a real object in motion does not need to do it in an instant of time. He can first measure its speed and then he can measure how long it takes for the object to pass by a particular location and then he can calculate its length.

But this thread is not about how an observer measures an object but rather how to correctly transform events from one frame to another in a meaningful way, so let's not confuse the two different ideas.
The only observers that matter when you are talking about events in a Frame of Reference are you and me because we are not in the Frame of Reference. If you ask me a question about events but you don't tell me the coordinates, then how am I supposed to know what you are talking about?

If you are talking about an observer in a scenario described by a Frame of Reference that you specify and you ask me what he will perceive, then you have to tell me where he is as a function of time, where the objects, clocks and other observers are as a function of time and then what you want to know about what he is observing. In other words, you have to give me the coordinates (or equations that let me calculate the coordinates) for all the events specifying everything in your scenario described by your FoR.

I can't read your mind. I have no idea what your concern is.

Remember, you started this thread with questions about transforming events from one FoR to another which has nothing to do with any observers anywhere, it's just math. Now you are asking about an observer, who must be described by a set of events if you want to use a FoR. As far as we are concerned, he is no different than any other mathematically described events, but you got to tell me what you have in mind.
 
  • #21
OK, my mobile device produced errors for your events but now that I can see them on a real computer I think what you are asking about is how an observer located at different distances from two events that occurred at the same time will perceive them. The observer has to wait for the images of the two events to propagate to him at the speed of light so he will see them occur at different times. If one was xa light-seconds away from him and the other one was xb light-seconds away (directions don't matter), then he will see the first one xa seconds after it happened and he will see the second one xb seconds after it happens. Therefore, if xa≠xb, then he will not see them simultaneously even though they happened simultaneously. Instead, he will see them separated in time by |xa-xb| seconds. We could use our FoR to determine the coordinates of the two events for the observer when he sees both events and calculate the difference in time, if we wanted to, and it would be real easy to do since everything is at rest in this particular FoR and so Proper Time for the observer is the same as Coordinate Time, but it will be the same answer that I just gave you, xa-xb.

But note, the observer has no awareness of the FoR that we used to describe this scenario, neither does he have any awareness of the coordinates for the two remote events, nor for the coordinates of his own perception events. We could have transformed the events from the FoR in which you described the scenario into any other FoR moving with respect to the first one, but then the observer's clocks will be time dilated and we have to do more work to calculate the difference in time that he perceives the two remote events. In other words, we have to take into account the Proper Time on his clock, not the Coordinate Times specified by the events.
 
  • #22
Suppose, you and me both at rest in our frame of reference. The two events happened simultaneously in our frame. You perceive it simultaneously, suppose at [itex]t_{g}[/itex], but I am not, suppose at [itex]t_{mf}[/itex] and [itex]t_{ms}[/itex]. Your proper time is same as my proper time. Your proper length is same as my proper length. Now You want to know [itex](t_{g1}, x_{g1}, y_{g1}, z_{g1})[/itex] and [itex](t_{g2}, x_{g2}, y_{g2}, z_{g2})[/itex]. And I want to know [itex](t_{m1}, x_{m1}, y_{m1}, z_{m1})[/itex] and [itex](t_{m2}, x_{m2}, y_{m2}, z_{m2})[/itex]. We both should agree on [itex](t_{g1}, x_{g1}, y_{g1}, z_{g1})=(t_{m1}, x_{m1}, y_{m1}, z_{m1})[/itex] and [itex](t_{g2}, x_{g2}, y_{g2}, z_{g2})=(t_{m2}, x_{m2}, y_{m2}, z_{m2})[/itex]. How can we find the co-ordinates?

If you are talking about that we can only transform one co-ordinates to another. Then how can I find the co-ordinates first. If I cannot then how can I conclude that the two simultaneous events not perceived simultaneously, but it is simultaneous in my frame?
 
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  • #23
Assuming that you and I have previously agreed on a common Frame of Reference in which we are both at rest, then no transformation is required and we both will have to wait for the signals coming from the remote events before we will know of their existences. Then we will have to use the techniques that astronomers use to determine distances and it could be very difficult or impossible.

But that is not why we talk about Frames of Reference. We talk about them for the sake of us who are not in the Frame of Reference so that we can meaningfully discuss various scenarios in an effort to understand Special Relativity. You see, we have the distinct advantage that we can be aware of everything that is happening without having to wait for the signals coming from the remote events.

Do you see the difference?
 
  • #24
ghwellsjr said:
You see, we have the distinct advantage that we can be aware of everything that is happening without having to wait for the signals coming from the remote events.

The same way can we not find two end points of the rod at same time? The rod is at rest in our frame.
 
  • #25
mananvpanchal said:
The same way can we not find two end points of the rod at same time? The rod is at rest in our frame.
OK, you want us to be in the scenario. I was trying to get you to see the difference between observers in a scenario and you and me sitting at our computers talking about observers in a scenario.

So, yes, let's pretend like we aren't sitting at our computers but we are in fact just in your scenario. Let me go back to your previous question first:
mananvpanchal said:
How can observer determine space components and time component of the events?
OK, we each take our rulers and I lay one end at the spatial point of the origin of our mutual Frame of Reference and point the other end in the x direction. You make a mark at the other end and call it x=1 and you place one end of your ruler against my ruler so that they are both pointing in the same direction. Then I pick up my ruler and go to the far end of your ruler and make a mark called x=2 and put one end of my ruler there and point the other end in the same direction as your ruler. Then you pick up your ruler and go to the end of my ruler and mark that point x=3. We repeat this process enough times to cover all our events along the +x direction. Then we both go back to our origin and we do the same thing in the opposite direction for the -x direction. Then we go back to our origin and mark out measured distances in the +y, -y, +z, and -z directions. After we get done with all of that, we fill in all the other locations so that any time an event occurs, we just look at the markings to see where it happened. Got it?

Now that we have established our grid for spatial locations, we have to put a clock at every one of those millions of points and we have to synchronize them to a master clock at the spatial origin according to Einstein's convention. These clocks are displaying coordinate time, one for each spatial coordinate and once we synchronize them, we don't allow them to move again. This is going to take a very long time so we better start with the master clock at a very large negative time so that when we get all done, we can each go back to the locations where we each belong when the scenario starts with the times on all the millions of coordinate clocks simultaneously reaching zero. From our respective locations, we can see all the clocks and we can see the markings describing their locations.

What will we see when the coordinate clocks at our locations finally reach zero? Well, all the ones that are one ruler length away will read -1, correct? And all the ones that are located two ruler lengths away will read -2 and so on, correct? This will be true individually for each of us but when I look at a given clock, I will see a different time on it than you will see at the same time. For example, suppose we are 1000 ruler lengths apart. When the clock at your location reads zero, you will see zero on it but I will see -1000 on it and I will see 0 on my clock but you will see -1000 on mine.

Now we're ready to start our scenario. Eventually the two events occur at two random locations but when their clocks have the same reading on them. Let's say the clocks are alarm clocks that flash a light when a particular time is reached and someone, unbeknownst to us has set two of them to go off when time 100 is reached on them. So we wait. Eventually all the clocks reach 100 and two of them emit the alarm. But we don't know this immediately because when we look at those two clocks (although we don't know yet which two they are), they will be reading numbers that are smaller by however far away they are from each of us. Eventually, the light from one of the clocks reaches one of us and that one notes where it was and what time was on it when the alarm went off. We do this three more times until each of us has determined the coordinates of the two events. We both of course will see the same pair of coordinates for the two events, but we will see them at different times according to the coordinate clocks at our locations.

Now your question:
mananvpanchal said:
The same way can we not find two end points of the rod at same time? The rod is at rest in our frame.
Well we just look at the coordinates that we have marked out previously for each of the two end points. We both of course will see the same location markings but at different times and maybe for different times at the two ends. I'm not sure what you mean by "at same time" since there are six clocks at six different clock locations which are all synchronized. Are you saying that our two coordinate clocks should be the same when we make our observations? Or are you saying that the two clocks at the two ends of the rod should be the same and that later on when we see those two clocks reading that time (which can be four different coordinate times for us) we each make note of the locations of those two clocks? But since the rod is not moving, the locations of its end points do not change with time, so why did you stipulate "at same time"?

Anyway, I hope this answers all your questions.
 
  • #26
Great description.

You understand what I am telling. But I don't know why this is going on.

All the points in the space has clocks and all the clocks is synchronized. Any two events happens simultaneously you perceive it as simultaneous, but I am not. But latter we find the actual time at when the events happened and we both agree on the time component and space components of the events. So, we both identify the events as simultaneous events. We both agree that time component is same for the both events. If we are wanted by some third one to transform the two events into his frame which is moving relative to us. We both uses the time components with which we both agreed, not that we perceived.

Now, take any two points in space and assume it as end points of rod. I measures the end points at different different time, but that is not the mean that time component of the both is not same. The clocks at the both point is still synchronized and showing the same reading. And the same as above I have to use the time component to transform which I have determined, not that I have perceived.
 
  • #27
mananvpanchal said:
Great description.

You understand what I am telling. But I don't know why this is going on.

All the points in the space has clocks and all the clocks is synchronized. Any two events happens simultaneously you perceive it as simultaneous, but I am not.
Now why do you say that? I thought you wanted both of us to be in the scenario at different locations but using the same Frame of Reference that we established together. We are in this same boat together and neither of us can perceive any two events simultaneously. Only those events that happen simultaneously equally distant from one of us will be perceived simultaneously. And other events that aren't simultaneous could be perceived as simultaneous by one or both of us. For example. Suppose you are 100 units away from the location of event P and 60 units away from the location of a event Q. Suppose the event P occurs at some time and then 40 units of time later, event Q occurs. After 60 more units of time, you will perceive both events at the same time, but looking at the coordinate clocks, you will see that P happened 100 units earlier and Q happened 40 units after that (or 60 units earlier than when you perceived them). On the other hand, if two events occurred at those two locations at the same time, you would perceive them as happening at different times (but the images on the coordinate clocks will tell you that they did happen simultaneously.
mananvpanchal said:
But latter we find the actual time at when the events happened and we both agree on the time component and space components of the events. So, we both identify the events as simultaneous events. We both agree that time component is same for the both events. If we are wanted by some third one to transform the two events into his frame which is moving relative to us. We both uses the time components with which we both agreed, not that we perceived.
That is correct.
mananvpanchal said:
Now, take any two points in space and assume it as end points of rod. I measures the end points at different different time, but that is not the mean that time component of the both is not same. The clocks at the both point is still synchronized and showing the same reading. And the same as above I have to use the time component to transform which I have determined, not that I have perceived.
But if the rod is stationary in the same frame in which we are stationary, then it won't matter what time you look at the coordinates, will it?
 
  • #28
ghwellsjr said:
But if the rod is stationary in the same frame in which we are stationary, then it won't matter what time you look at the coordinates, will it?

That is what I was trying to say
mananvpanchal said:
You can have only one time component for all space components of rod if rod is at rest in your frame.
 
  • #29
The clocks at both ends of rod is synchronized. If clocks emits light pulse at every second. Then suppose, both clocks emits light pulse simultaneously at [itex]t_0[/itex]. The both clocks is at different distance from us. So we perceive first pulse at [itex]t_1[/itex] and second at [itex]t_2[/itex]. But we know the location of clocks, so we can conclude that we perceive both pulse unsimultaneously yet the events happened simultaneously. We can calculate that both events occurred at [itex]t_0[/itex].

If someone tell us to transform to another frame which is moving relative to us, then we use [itex]t_0[/itex] as time component of the events, not [itex]t_1[/itex] and [itex]t_2[/itex]. So, we can say that we have one time component for all space component of the rod if the rod is steady in our frame.
 
  • #30
mananvpanchal said:
The clocks at both ends of rod is synchronized. If clocks emits light pulse at every second. Then suppose, both clocks emits light pulse simultaneously at [itex]t_0[/itex]. The both clocks is at different distance from us. So we perceive first pulse at [itex]t_1[/itex] and second at [itex]t_2[/itex]. But we know the location of clocks, so we can conclude that we perceive both pulse unsimultaneously yet the events happened simultaneously. We can calculate that both events occurred at [itex]t_0[/itex].
Yes, exactly right.
mananvpanchal said:
If someone tell us to transform to another frame which is moving relative to us, then we use [itex]t_0[/itex] as time component of the events, not [itex]t_1[/itex] and [itex]t_2[/itex]. So, we can say that we have one time component for all space component of the rod if the rod is steady in our frame.
No, you don't want to use any of those time components.

First off, when you are doing transforms from one frame to another, the t1 and t2 times are not relevant for the events concerning the ends of the rod because the observers are at a different location than either end of the rod. It won't even matter if you or I or any other observer exists in the scenario. Just consider only two events at the ends of the rods. Now since the rod is not moving in the initial FoR, it doesn't matter what the time coordinates are for the two events that we want to use to determine the end points of the rod, will it? We can use any pair of events with different time components we want in the rest frame of the rod and as long as one event is located at one end of the rod and the other event is located at the other end of the rod, the difference between the spatial components of those two events will give us the Proper Length (or the Rest Length) of the rod, correct?

So the goal is to find two events, one at either end of the rod (and there are an infinite number of possibilities) such that when you transform them into another frame, they end up having the same time component. Do you understand what I'm saying and why I'm saying it?

If you understand, then the problem becomes one of how you find two events that qualify and as I said earlier, there are many different ways to do this as I outlined in posts #2, #9 and the end of #16, as well as what others have been telling you.
 
  • #31
mananvpanchal in this thread like in other threads you are making assumptions that you know how things should work, yet they don't work the way you think they do. Go back and look at what people posted with an open mind. I feel you misunderstand much of what people say due to you trying to convert what they say to what you think they say.
 
  • #32
mananvpanchal said:
If someone tell us to transform to another frame which is moving relative to us, then we use [itex]t_0[/itex] as time component of the events, not [itex]t_1[/itex] and [itex]t_2[/itex]. So, we can say that we have one time component for all space component of the rod if the rod is steady in our frame.

Why do those two clocks being in sync in one frame of reference mean that they need to be in sync in another?
 
  • #33
darkhorror said:
Why do those two clocks being in sync in one frame of reference mean that they need to be in sync in another?
It's very important to realize that the two clocks that are in sync at the two ends of the rod in the rest frame of the rod are coordinate clocks for that frame. In another frame moving with respect to the first rest frame, we don't pay attention to those two clocks anymore. Rather, we have a whole new set of coordinate clocks that have been synchronized together with each other to define the new frame. Now, there are no longer just two new coordinate clocks at the two ends of the rod. Instead, we have a whole bunch of new clocks that are moving with respect to the rod, or we could say the rod is moving with respect to the new set of coordinate clocks. That's why, if we looked at the two coordinate clocks in the new frame when the coordinate clocks in the original frame had the same time on them, the new clocks would have different times on them and their spacing apart is different and we end up with a meaningless result. Instead, we have to look at a new coordinate clock located at one end of the rod and wait until the other end of the rod moves past a new clock with the same time on it and then see how far apart those two new clocks are which tells us the contracted length of the rod in the new frame.
 
  • #34
Ok, I accept it that we have to pick two different time components in rest frame to find same time component in moving frame, and with this we can get now length contraction.

Now, please take a look in below image.

lengh contraction 02.JPG


This is same as displayed in the link http://en.wikipedia.org/wiki/Length_contraction#Geometrical_representation.

We can say that OA' is length between two events at rest in S'. When we transform O and A', we get O and A in S frame. We can easily say that OA' > OA. So for moving frame S, the length is contracted.

Now, take two events happened at rest in S' frame at O and B' time. When we transform the two events, we get O and B in S frame. We can again easily say that OB' > OB. So, we have to say that time elapsed between two events in S' frame is more than S frame. So time is running faster in moving frame than rest frame.
 
  • #35
darkhorror said:
mananvpanchal in this thread like in other threads you are making assumptions that you know how things should work, yet they don't work the way you think they do. Go back and look at what people posted with an open mind. I feel you misunderstand much of what people say due to you trying to convert what they say to what you think they say.

Can you explain me why I am wrong?
 

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